64. Minimum Path Sum

• 題目

  • Given a m x n grid filled with non-negative numbers, find a path from top left to bottom right which minimizes the sum of all numbers along its path.

  • Note: You can only move either down or right at any point in time.

  • Example:

    Input:
    [
      [1,3,1],
      [1,5,1],
      [4,2,1]
    ]
    Output: 7
    Explanation: Because the path 1→3→1→1→1 minimizes the sum.
     
    

• 解題思路

  • 本題比較簡單，題意是尋找m x n格中左上角到右下角路徑的最小距離，並且沒有負權值
  • dp[i][j]:起點到點 [i,j]的最小距離
  • 因為沒有負路徑和尋路方向只能向下或者向右，因此對於每個點(除起點)，其dp[i][j]的表示為 dp[i][j] = min(dp[i-1][j],dp[i][j-1]) + grid[i][j]

• 實現程式碼

  class Solution {
  public:
      int minPathSum(vector<vector<int> >&
   
   grid) {
          int m = grid.size();
          int n = grid[0].size();
          vector<vector<int> > dp(m, vector<int>(n, 0));
          dp[0][0] = grid[0][0];
          for(int i = 0; i < m; ++i)
          	for(int j = 0; j < n ; ++j) {
          		if(i == 0 && j == 0)    continue;
                          if
   
  (i == 0)
          			dp[i][j] = dp[i][j-1] + grid[i][j];
          		else if(j == 0)
          			dp[i][j] = dp[i-1][j] + grid[i][j];
          		else
          			dp[i][j] = min(dp[i-1][j], dp[i][j-1]) + grid[i][j];
          	}
  
          return dp[m-1][n-1];
      }
  };